# Dividing polynomial

• Mar 22nd 2011, 10:04 AM
Devi09
Dividing polynomial
I'm having problems dividing the polynomial 2x^5 + 2x^3 + x^2 + 1 by x^2 + 1
• Mar 22nd 2011, 10:19 AM
HallsofIvy
Looks straight forward to me. But I don't know what to tell you without seeing what you did. Did you remember to include "0x" in the divisor and "\$\displaystyle 0x^4\$" and "\$\displaystyle 0x\$" in the dividend? How many times does \$\displaystyle x^2\$ divide into \$\displaystyle 2x^5\$?
• Mar 22nd 2011, 10:39 AM
Devi09
It goes in 2x^3 times but I get x^2 as the next quotient part but the correct quotient should be 2x^3 - 1 according to the answer the book gives
• Mar 22nd 2011, 03:03 PM
Devi09
\$\displaystyle x^2+0x+1sqrt(2x^5+0x^4+2x^3+x^2+0x+1)\$

This is the subtracting parts: The quotient i got is \$\displaystyle 2x^3+x^2+x+1\$

Part 1 \$\displaystyle 2x^5+0x^4+2x^3\$

Part 2 \$\displaystyle x^4+x^2\$

\$\displaystyle x^4+x^3+x^2\$

Part 3 \$\displaystyle x^3+0x\$

\$\displaystyle x^3+x^2+x\$

Part 4 \$\displaystyle x^2+1\$

\$\displaystyle x^2+x+1\$

Final \$\displaystyle x\$
• Mar 23rd 2011, 04:54 AM
HallsofIvy
Yes, \$\displaystyle x^2\$ divides into \$\displaystyle 2x^5\$ \$\displaystyle 2x^3\$ times. And multiplying the entire divisor by \$\displaystyle 2x^3\$ gives \$\displaystyle 2x^5+ 2x^3\$
Now subtract that from the dividend:
\$\displaystyle 2x^5+ 2x^3+ x^2+ 1- (2x^5+ 2x^3)= x^2+ 1\$.

And now, of course, \$\displaystyle x^2+ 1\$ divides into that exactly once: the quotient is \$\displaystyle 2x^3+ 1\$.